Amoebas, Monge-Amp`Ere Measures, and Triangulations of the Newton

AMOEBAS, MONGE-AMPERE` MEASURES, AND TRIANGULATIONS OF THE NEWTON POLYTOPE

MIKAEL PASSARE and HANS RULLGARD˚

Abstract The amoeba of a holomorphic function f is, by deﬁnition, the image in Rn of the zero locus of f under the simple mapping that takes each coordinate to the logarithm of its modulus. The terminology was introduced in the 1990s by the famous (biologist and) mathematician Israel Gelfand and his coauthors Kapranov and Zelevinsky (GKZ). In this paper we study a natural convex potential function N f with the property that its Monge-Ampere` mass is concentrated to the amoeba of f . We obtain results of two kinds; by approximating N f with a piecewise linear function, we get striking combinatorial information regarding the amoeba and the Newton polytope of f ; by computing the Monge-Ampere` measure, we ﬁnd sharp bounds for the area of amoebas 2 in R . We also consider systems of functions f1,..., fn and prove a local version of the classical Bernstein theorem on the number of roots of systems of algebraic equations.

1. Introduction The classical Jensen formula can be regarded as a relation between the zeros of a holomorphic function f and the properties of a certain convex function associated to f . In this paper we consider an analogue of this convex function, which we denote N f , in the case where f is a function of several variables, and we search for relations with the zero locus of f . The results we find are of two kinds. In Section 3, we show that approximation of N f by a piecewise linear function leads to an approximation of the so-called amoeba by a polyhedral complex (Th. 1). When f is a polynomial, this polyhedral complex is dual (in a precise sense) to a certain subdivision of the Newton polytope of f . An explicit computation of the polyhedral complex amounts to evaluating certain parameters depending on the coefficients of f . It turns out that these satisfy a GKZ-type system of differential equations and can be expanded in a hypergeometric power series involving the coefficients (Th. 3).

DUKE MATHEMATICAL JOURNAL Vol. 121, No. 3, c 2004 Received 18 September 2002. Revision received 3 April 2003. 2000 Mathematics Subject Classiﬁcation. Primary 32A60; Secondary 52A41, 52B20.

481 482 PASSARE and RULLGARD˚

In Section 5 we investigate the measures arising when Monge-Ampere` or Laplace operators act on N f . We ﬁnd several relations between such measures and the hypersurface deﬁned by f (Ths. 5 and 6). The former can be regarded as a local variant of the Bernstein formula for the number of solutions to a system of polynomial equations. We also obtain an estimate on the Monge-Ampere` measure, which gives an upper bound on the area of the amoeba of a polynomial in two variables (Th. 7 and Cor. 1). This estimate was used in [8] to give a new characterization of so-called Har- nack curves, which are fundamental for Hilbert’s sixteenth problem in real algebraic geometry.

2. Background Suppose that f is an entire function in the complex plane with zeros a1, a2, a3,... ordered so that |a1| ≤ |a2| ≤ · · · . Assume for simplicity that f (0) 6= 0. The Jensen formula states that

m 1 Z 2π X r log | f (reit )| dt = log | f (0)| + log , 2π |ak| 0 k=1 where m is the largest index such that |am| < r. If this expression is considered as a function N f of log r, there is a strong connection between this function and the zeros of f . Thus it follows immediately that N f is a piecewise linear convex function whose gradient is equal to the number of zeros of f inside the disc {|z| < r}. The second derivative of N f , in the sense of distributions, is a sum of point masses at log |ak|, k = 1, 2, 3,.... In this paper we consider a certain generalization of the function occurring in the Jensen formula to holomorphic functions of several variables. Let be a convex open set in Rn, and let f be a holomorphic function de- −1 n n ﬁned in Log (), where Log : (C \{0}) −→ R is the mapping (z1,..., zn) 7→ (log |z1|,..., log |zn|). In [12] Ronkin considers the function N f deﬁned in by the integral 1 Z log | f (z ,..., z )| dz ··· dz = 1 n 1 n N f (x) n . (1) (2πi) Log−1(x) z1 ··· zn

As we will see, the function N f retains some of its properties from the one- variable case, while others are lost or attain a new form. For example, N f is a convex function, but it is no longer piecewise linear. In Section 3 we consider the conse- quences of approximating N f by a piecewise linear function. In Section 5 we investi- −1 gate the relation between local properties of N f and the hypersurface f (0). The properties of the function N f are closely related to the amoeba of f . The amoeba of f , which we denote A f , is deﬁned to be the image in of the hypersurface f −1(0) under the map Log. The term amoeba was ﬁrst used by Gelfand, Kapranov, and Zelevinsky [5] in the case where f is a polynomial. AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 483

Suppose that E is a connected component of the amoeba complement \ A f . It is not difﬁcult to show that all such components are convex. For example, Log−1(E) is the intersection of Log−1() with the domain of convergence of a certain Laurent series expansion of 1/f , and domains of convergence of Laurent series are always logarithmically convex. In [4] Forsberg, Passare, and Tsikh deﬁned the order of such a component to be the vector ν = (ν1, . . . , νn) given by the formula

1 Z ∂ f z dz ··· dz = j 1 n ∈ ν j n , x E. (2) (2πi) Log−1(x) ∂z j f (z)z1 ··· zn Here x may be any point in E. They proved, for the case when f is a Laurent polynomial, that the order is an integer vector, that is, ν ∈ Zn, that ν is in the Newton polytope of f , and that two distinct components always have different orders. These conclusions remain true, with essentially the same proofs, in the more general setting considered here. Ronkin proved in [12] a theorem that in the language of amoebas amounts to the following statement.

THEOREM Let f be a holomorphic function as above. Then N f is a convex function. If U ⊂ is a connected open set, then the restriction of N f to U is afﬁne linear if and only if U does not intersect the amoeba of f . If x is in the complement of the amoeba, then grad N f (x) is equal to the order of the complement component containing x.

Sketch of proof The convexity of N f follows from a general theorem because log | f | is plurisubharmonic (see, e.g., [11, Cor. 1, p. 84]). Differentiation with respect to x j under the integral sign in definition (1) of N f yields precisely the real part of the integral (2) defining the order. However, the integral (2) is always real valued, and this shows immediately that N f is affine linear in each connected component of \ A f . The fact that N f is not linear on any open set intersecting the amoeba of f can be proved in several ways. It follows, for instance, from our results in Section 5.

3. Triangulations and polyhedral subdivisions In his doctoral thesis [3], Mikael Forsberg noted that the amoebas of many polynomials in two variables have the appearance of slightly thickened graphs. Moreover, there seemed to be some kind of duality between these graphs and certain subdivisions (of- ten triangulations) of the Newton polygon of the polynomial. We make this precise by associating to an amoeba A f a polyhedral complex, which we call its spine. This is done as follows. 484 PASSARE and RULLGARD˚

Assume that f is a holomorphic function deﬁned in (C \{0})n, and write

n n A = α ∈ Z ; R \ A f has a component of order α .

It may happen that A is empty, but we assume that this is not the case. The most interesting situation arises when A has plenty of points, for example, when the convex hull of A coincides with the Newton polyhedron of f . This always happens if f is a Laurent polynomial. For α ∈ A, let cα = N f (x) − hα, xi, (3) where x is any point in the complement component of order α, and let S(x) = max cα + hα, xi . (4) α∈A This S(x) is a piecewise linear function approximating the Ronkin function. The spine S f is deﬁned as the corner set of S, that is, the set of x where S(x) is nonsmooth.

Figure 1. Amoeba of the polynomial + 5 + 2 + 3 2 + 3 4 1 z1 80z1z2 40z1z2 z1z2 (shaded) together with its spine (solid) and the dual triangulation of the Newton polytope

We now deﬁne a precise meaning to the statement that the spine is dual to a certain subdivision of the convex hull of A. For an example, see Figure 1.

Definition 1 Let K be a convex set in Rn. A collection T of nonempty closed convex subsets of K is called a convex subdivision if it satisfies the following conditions. (i) The union of all sets in T is equal to K . (ii) If σ, τ ∈ T and σ ∩ τ is nonempty, then σ ∩ τ ∈ T . (iii) If σ ∈ T and τ is any subset of σ , then τ ∈ T if and only if τ is a face of σ. We say that T is locally finite if every compact set in K intersects only a finite number of σ ∈ T , and we say that T is polytopal if every σ ∈ T is a polytope. AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 485

Here a face of a convex set σ means a set of the form {x ∈ σ ; hξ, xi = supy∈σ hξ, yi} for some ξ ∈ Rn. If σ, τ ∈ T , and τ is a proper subset of σ, then it follows from condition (iii) that dim τ < dim σ. Hence any descending chain σ1 ⊃ σ2 ⊃ · · · in T stabilizes after a ﬁnite number of steps. Therefore the intersection of any collection of sets in T is also in T . If τ ⊂ σ are convex sets, we deﬁne

cone(τ, σ ) = t(x − y); x ∈ σ, y ∈ τ, t ≥ 0 .

Clearly, this set is a convex cone. If C is a convex cone, its dual is deﬁned to be the cone C∨ = {ξ ∈ Rn; hξ, xi ≤ 0, ∀x ∈ C}. If C is closed, it is well known that C∨∨ = C.

Deﬁnition 2 Let K, K 0 be convex sets in Rn, and let T, T 0 be convex subdivisions of K, K 0. We say that T and T 0 are dual (to each other) if there exists a bijective map T → T 0, denoted σ 7→ σ ∗, satisfying the following conditions. (i) For σ, τ ∈ T , τ ⊂ σ if and only if σ ∗ ⊂ τ ∗. (ii) If τ ⊂ σ, then cone(τ, σ ) is dual to cone(σ ∗, τ ∗).

Notice that cone(σ, σ ) is the affine subspace spanned by σ . Hence the second condition implies that σ is orthogonal to σ ∗ and, in particular, that dim σ + dim σ ∗ = n. Next, we show that a convex, piecewise linear function determines a convex subdivision and that its Legendre transform determines a dual subdivision. Let S(x) be n a function of the form (4), where A is now any discrete subset of R and where cα are arbitrary numbers such that S(x) is finite for all x. The Legendre transform of S is defined by S˜(ξ) = sup hξ, xi − S(x), x∈Rn and it is again a piecewise linear convex function with finite values defined in the convex hull K of A. Consider the function

P(ξ, x) = S(x) + S˜(ξ) − hξ, xi (5) deﬁned on K × Rn.

LEMMA 1 The function P(ξ, x) deﬁned by (5) has the following properties. (i) P(ξ, x) ≥ 0. 486 PASSARE and RULLGARD˚

(ii) P(ξ, x) is convex in each argument when the other is held ﬁxed. (iii) For x, y ∈ Rn and ξ, η ∈ K , hξ −η, x − yi = −P(ξ, x)+ P(ξ, y)+ P(η, x)− P(η, y). (iv) For every x ∈ Rn there is a ξ ∈ K such that P(ξ, x) = 0, and for every ξ ∈ K there is an x ∈ Rn such that P(ξ, x) = 0. n (v) If x, y ∈ R , there is a t0 > 0 such that P(ξ, x + ty) is a linear function of t n for t ∈ [0, t0]. If ξ ∈ K and η ∈ R is such that ξ + tη ∈ K for small positive t, then there is a t0 > 0 such that P(ξ + tη, x) is a linear function of t for t ∈ [0, t0].

Proof Properties (i), (ii), and (iii) are trivial. n (iv) If x ∈ R , there is an α ∈ A such that S(x) = cα + hα, xi. This implies that ˜ S(α) = −cα, so P(α, x) = 0. Let ξ ∈ K be given. Then we must show that the function S(x)−hξ, xi attains its ˜ smallest value at some point x0. Indeed, if this is the case, then S(ξ) = hξ, x0i− S(x0) which means that P(ξ, x0) = 0. To show that S(x) − hξ, xi attains its smallest value, assume first that ξ is in the interior of K . Then S(x) − hξ, xi → +∞ when x → ∞, and it follows that the function attains its smallest value. The same argument works if dim K < n and ξ is in the relative interior of K . So, suppose that ξ is on the boundary of K , take y ∈ Rn such that ξ is in the relative interior of the face L = {η ∈ K ; hη, yi = supα∈Ahα, yi}, and let B = A∩L. Define SB(x) = maxα∈B(cα +hα, xi). By the preceding argument, SB(x) − hξ, xi attains its infimum at some point x0 and hence at x0 + ty for every t ∈ R. Since S(x) is finite for all x, there can be only a finite number of α ∈ ArB such that cα +hα, x0i > SB(x0). Moreover, for each α ∈ A r B, SB(x0 +ty)−hα, x0 +tyi is an increasing function of t which tends to +∞ when t → +∞. This implies that SB(x0 + ty) = S(x0 + ty) for sufficiently large t. Since S(x) ≥ SB(x), it follows that S(x) − hξ, xi attains its smallest value at x = x0 + ty for sufficiently large t. n (v) Let x ∈ R , let A1 = {α ∈ A; cα + hα, xi = S(x)} and A2 = A r A1, and let S j (y) = maxα∈A j (cα + hα, yi) for j = 1, 2. Then S(x) = S1(x) > S2(x), so it follows that S(y) = S1(y) for y in a neighborhood of x. Moreover, S1(x + ty) = + h i S(x) t maxα∈A1 α, y is linear in t for all t > 0. This proves the first part. To prove the second part, let ξ ∈ K , let η ∈ Rn, let σ = {x ∈ Rn; P(ξ, x) = 0}, and take x ∈ σ with hη, xi as large as possible. We claim that S˜(ξ + tη) = S˜(ξ) + thη, xi for small positive t. ˜ ˜ To prove this claim, note that S(ξ +tη) ≥ hξ +tη, x0i− S(x0) = S(ξ)+thη, x0i. On the other hand, if we write S(x) = max(S1(x), S2(x)) as in the first part, then ˜ ˜ ˜ S ≥ S1, so S(ξ + tη) ≤ S1(ξ + tη) = S(ξ) + thη, x0i for small positive t. AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 487

0 Deﬁne T to be the collection of all sets σξ = {x; P(ξ, x) = 0} for ξ ∈ K . Deﬁne T 0 0 analogously as the collection of all sets σx = {ξ ∈ K ; P(ξ, x) = 0}. Then T and T are dual convex subdivisions.

PROPOSITION 1 With notation as in the preceding paragraph, T and T 0 are dual convex subdivisions of n 0 ∗ T 0 R and K , where the correspondence between T and T is given by σ = x∈σ σx = 0∗ T n {ξ ∈ K ; P(ξ, x) = 0, ∀x ∈ σ} and σ = ξ∈σ 0 σξ = {x ∈ R ; P(ξ, x) = 0, ∀ξ ∈ σ 0} for σ ∈ T , σ 0 ∈ T 0. Moreover, T is locally ﬁnite and T 0 is polytopal.

Proof 0 0 First, we must check that T and T are convex subdivisions. The cells σξ and σx are convex by properties (i) and (ii) of Lemma 1, and they are nonempty by property (iv). S n S 0 It also follows from property (iv) that ξ∈K σξ = R and that x∈Rn σx = K . Hence condition (i) of Definition 1 is satisfied. ∩ ∩ = ∈ ∩ If σξ1 σξ2 is nonempty, we claim that σξ1 σξ2 σ(ξ1+ξ2)/2. Indeed, if x σξ1 = = σξ2 , then P(ξ1, x) P(ξ2, x) 0. Since P(ξ, x) is nonnegative and convex in ξ, it + = ∈ ∈ follows that P((ξ1 ξ2)/2, x) 0, so x σ(ξ1+ξ2)/2. Conversely, let x σ(ξ1+ξ2)/2, ∈ ∩ = = + = and let x0 σξ1 σξ2 . Then P(ξ1, x0) P(ξ2, x0) P((ξ1 ξ2)/2, x) 0, and by the preceding argument, P((ξ1 + ξ2)/2, x0) = 0. By property (iii) of Lemma 1, with y = x0, ξ = (ξ1 + ξ2)/2, and η = ξ1 or ξ2, it follows that P(ξ1, x) = hξ1 − ξ2, x − x0i/2 = −P(ξ2, x). Since P is nonnegative, it follows that P(ξ1, x) = P(ξ2, x) = 0, ∈ ∩ so x σξ1 σξ2 . This verifies condition (ii) of Definition 1 for T . ⊂ = − Suppose that σξ1 σξ2 . We must show that σξ1 is a face of σξ2 . Let η ξ1 ξ2, and let τ = {x ∈ σξ ; hη, xi = sup ∈ hη, yi} be the corresponding face of σξ . We 2 y σξ2 2 = ∈ ∈ claim that σξ1 τ. Indeed, if x σξ1 and y σξ2 , it follows from property (iii) of h − i = ≥ ⊂ Lemma 1 that η, x y P(ξ1, y) 0, which shows that σξ1 τ. On the other ∈ ∈ ⊂ = h − i = − ∈ hand, if x τ and y σξ1 τ, then 0 x y, η P(ξ1, x), so x σξ1 , = completing the proof that σξ1 τ. Conversely, let ξ2 ∈ K , and let τ = {x ∈ σξ ; hη, xi = sup ∈ hη, yi} be a face 2 y σξ2 ∈ = + of σξ2 . We must find ξ1 K such that τ σξ1 . Take t0 > 0 so that P(ξ2 tη, x) ∈ [ ] = + ∈ is linear for t 0, t0 , and let ξ1 ξ2 (t0/2)η. For all x / σξ2 , P(ξ2, x) > 0 and + ≥ ∈ ∈ + = h − i P(ξ2 t0η, x) 0. For all x σξ2 r τ and y τ, P(ξ2 t0η, x) t0 η, y x > 0 and P(ξ2, x) = 0. From the linearity of P, it follows that P(ξ1, x) > 0 for all x ∈/ τ. If x, y ∈ τ, then P(ξ1, x) − P(ξ1, y) = P(ξ2, x) − P(ξ2, y) − (t0/2)hη, x − yi = 0. = = Since P(ξ1, x) 0 for some x, it follows that σξ1 τ. This completes the proof that T is a convex subdivision. The proof for T 0 is analogous. Next, we check the duality property. It is clear that σ ∗ is either empty or a cell 488 PASSARE and RULLGARD˚

0 ⊂ ⇒ ∗ ⊂ ∗ ∈ ∗ ⊂ ∗∗ of T , that τ σ σ τ , that ξ σξ , and that σ σ . Together with the corresponding statements for cells in T 0, this implies that σ 7→ σ ∗ is an inclusion reversing bijection from the cells of T to the cells of T 0. Thus it remains to check condition (ii) of Deﬁnition 2. Let τ ⊂ σ be cells in T . If x − y ∈ cone(τ, σ ) and η − ξ ∈ cone(σ ∗, τ ∗), where x ∈ σ, y ∈ τ, ξ ∈ σ ∗, η ∈ τ ∗, it follows from property (iii) of Lemma 1 that hη − ξ, x − yi = −P(η, x) ≤ 0, which implies that cone(σ ∗, τ ∗) ⊂ cone(τ, σ )∨. ∨ To prove the opposite inclusion, assume σ = σξ , and let η ∈ cone(τ, σ ) . Take ∗ t0 > 0 such that P(ξ + tη, x) is linear for t ∈ [0, t0]. We claim that ξ + (t0/2)η ∈ τ . Since ξ ∈ σ ∗, this implies that η ∈ cone(σ ∗, τ ∗), as desired. To prove the claim, note that for all x ∈/ σ, P(ξ, x) > 0 and P(ξ + t0η, x) ≥ 0, so from the linearity of P, it follows that P(ξ + (t0/2)η, x) > 0. From property (iv) of Lemma 1, it follows that P(ξ + (t0/2)η, x) must vanish for some x ∈ σ. But if x ∈ σ and y ∈ τ, it follows from the assumption η ∈ cone(τ, σ )∨ that t t t 0 ≥ 0 hη, x − yi = P ξ + 0 η, y − P ξ + 0 η, x , 2 2 2 so P(ξ +(t0/2)η, y) ≤ P(ξ +(t0/2)η, x). Hence P(ξ +(t0/2)η, y) = 0 for all y ∈ τ, ∗ which means that ξ + (t0/2)η ∈ τ , as was claimed.

Now we specialize to the situation considered at the beginning of this section. Thus f is an entire holomorphic function, A is the set of orders of complement components of A f , which we assume is nonempty, and S(x) is the piecewise linear approximation of the Ronkin function deﬁned by (3) and (4).

THEOREM 1 Let f and A be as above. Then there exist a convex subdivision T of Rn and a dual convex subdivision T 0 of the convex hull of A with the following properties. (i) The spine S f is the union of all cells in T of dimension less than n, and it is a subset of A f . (ii) For α ∈ A, the cell of T dual to the point {α} ∈ T 0 contains the complement component of order α. (iii) If the convex hull of A coincides with the Newton polyhedron of f (e.g., if f is a Laurent polynomial), then the spine is a deformation retract of the amoeba.

Proof The dual subdivisions T and T 0 are constructed from S and its Legendre transform according to Proposition 1. Let Eα denote the complement component of order α, and let Fα = {x; S(x) = cα + hα, xi}. Then Eα ⊂ Fα, and by definition, the spine is the union of the boundaries of the sets Fα. From this it follows that S f ⊂ A f . It AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 489 is readily verified that the Fα are the maximal cells of T and that each Fα is dual to {α} ∈ T 0. This proves (ii), and it also follows that the cells of T of dimension less than n are precisely the proper faces of the Fα. The union of all such cells is the union of the boundaries of the Fα, that is, the spine. This proves (i). To construct the deformation retraction in (iii), take a point pα in each boundary component Eα, and consider the collection of all line segments from pα to the boundary of Fα. We claim that the union of all these segments contains the amoeba. Therefore we can deform the amoeba along the segments up to the spine. To prove the claim, consider one of these points pα, and consider a ray γ = {pα + tv; t ≥ 0} which does not intersect the boundary of Fα. We must show that this ray does not intersect the amoeba. The assumption that γ does not intersect the boundary of Fα implies that hα, vi ≥ hβ, vi for every β ∈ A and hence for every β in the Newton polyhedron of f . Let r > 0 be any real number, and let Br be the ball of radius r centered at pα. We show that γ ∩ Br does not intersect the amoeba for any ray γ not intersecting the boundary of Fα. Let > 0 be a small number, to be specified in a moment, and write f = f1 + f2, where f1 is a Laurent polynomial whose Newton polytope is contained in −1 the Newton polytope of f , such that | f2(z)| < | f1(w)| for all z ∈ Log (Br ) and −1 w ∈ Log (pα). Let C be the set of exponent vectors occurring in f1, and let s be an integer vector such that hα, si ≥ hβ, si for all β ∈ C. −s −s Consider the function φ(ζ ) = f (a1ζ 1 ,..., anζ n ) of one complex variable hα,si ζ. Then ζ φ(ζ ) is holomorphic in the unit disc. Since pα ∈ Eα, the number of poles minus the number of zeros in the unit disc of φ is hα, si; hence ζ hα,siφ(ζ ) does not vanish in the unit disc. By applying the maximum principle to its reciprocal and −1 letting a vary over Log (pα), this implies that

min | f (z)| ≥ ehα,tsi min | f (z)|. −1 −1 Log (pα+ts) Log (pα) By continuity, this inequality remains true if we drop the requirement that s be an integer vector, and, in particular, it holds for s = v. Now it follows that pα + tv∈ / A f for |tv| < r, provided that < e−r|α|.

Remark. Suppose that f is a polynomial. In this case, the convex hull of A coincides with the Newton polytope of f . The convex subdivision of the Newton polytope constructed in Theorem 1 is an example of the coherent triangulations that play an important role in the theory of discriminants (see [5]). 490 PASSARE and RULLGARD˚

4. Dependence of the spine on the polynomial In this section we restrict attention to Laurent polynomials f . The spine S f of its amoeba is easy to compute once we know the coefficients cα from (3). We study here how the cα depend on the coefficients of the polynomial f . The main result is an expansion of the coefficients cα as a hypergeometric series in the coefficients of the polynomial. We also show that the coefficient cα depends only on the truncation of f to the smallest face of its Newton polytope containing α, and we work out a few examples. Consider a Laurent polynomial

X α f (z) = wαz α∈C n with coefficients wα, where C is a finite subset of the lattice Z . To study the dependence of cα on the coefficients, it is useful to introduce the functions 1 Z log( f (z)/zα) dz ··· dz = 1 n 8α(w) n , (6) (2πi) Log−1(x) z1 ··· zn where x is in the component of order α. This means that

cα = Re 8α.

Notice that 8α is a holomorphic function in the coefﬁcients w with values in C/2πiZ, deﬁned whenever the complement of the amoeba of f has a component of order α.

Remark. The functions 8α, or rather ∂8α/∂wα, were used in [2] in the study of constant terms in powers of Laurent polynomials. The main result was obtained by showing, essentially, that the second term in representation (9) is nonconstant along every complex line parallel to the wα-axis, provided that α is not a vertex of the Newton polytope of f . These functions were also used in [13] to prove the existence of amoebas with prescribed complement components.

Example Consider a one-variable polynomial f (z) = (z + a1) ··· (z + aN ) = w0 + · · · + N−1 N wN−1z + z . Let 0 ≤ k ≤ N, and assume that |a1| ≤ · · · ≤ |ak| < r < |ak+1| ≤ AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 491

· · · ≤ |aN |. Then one ﬁnds that

In this case, we observe that 8k(w) may be continued as a ﬁnitely branched holomorphic function whose branches correspond to various permutations of the roots of f . Incidentally, the sum of all branches of exp 8k(w) is equal to wk. For polynomials of several variables, the analytic continuation of exp 8α(w) is in general inﬁnitely branched.

Let 0 be a face of the Newton polytope of f , and let f |0 denote the truncation of f to 0; that is, let X α f |0(z) = wαz . α∈0 It is known that when α ∈ 0 ∩ Zn, the complement of the amoeba of f has a component of order α precisely if the complement of the amoeba of f |0 does (see [4, Prop. 2.6] and also [5]).

THEOREM 2 Let f be a Laurent polynomial, and let 0 be a face of the Newton polytope of f . If α ∈ 0 and the complement of A f has a component of order α, then 8α( f ) = 8α( f |0). In particular, if α is a vertex of the Newton polytope of f , then 8α(w) = log wα.

Proof Take an outward normal v to the Newton polytope at 0. If α ∈ 0 and x is in the n component of R \ A f of order α, it is known that x + tv is also in that component for all t > 0. Therefore 1 Z f (z) dz ··· dz − | = 1 n 8α( f ) 8α( f 0) n log , (2πi) Log−1(x+tv) f |0(z) z1 ··· zn and here the integrand clearly tends to zero when t → ∞.

The main result of this section is that the functions 8α very nearly satisfy a system of GKZ hypergeometric equations. Let A be a finite set in Zn+1 which is assumed to lie 492 PASSARE and RULLGARD˚ in the affine sublattice defined by setting the first coordinate equal to 1. Denote by CA the set of all tuples (wα; α ∈ A, wα ∈ C). A GKZ system associated to A is a system of differential equations of the form

u v A X (∂ − ∂ )8 = 0 if u, v ∈ N , (uα − vα)α = 0 (7) α∈A and X αwα∂α8 = λ8. (8) α∈A u Here ∂α = ∂/∂wα, and ∂ is the obvious multi-index notation for a higher partial derivative. The constant λ denotes an arbitrary vector in Cn+1. The system of differential equations in the following theorem is very similar to a GKZ system where A = {(1, γ ); γ ∈ C} and λ = 0. The only difference is that we introduce a nonhomogeneous term in equation (8). It is easy to see that all partial derivatives of 8α, which are actually coefﬁcients in a Laurent series expansion of 1/f , satisfy a true GKZ hypergeometric system.

THEOREM 3 P γ Let f (z) = γ ∈C wγ z be a Laurent polynomial with C being a ﬁxed ﬁnite subset n of Z . Then the holomorphic functions 8α have the power series expansion (−k − 1)! X α kα−1 k 8α(w) = log wα + Q (−1) w , (9) β6=α kβ ! k∈Kα where

n C X X o Kα = k ∈ Z ; kα < 0, kβ ≥ 0 if β 6= α, kγ = 0, γ kγ = 0 . (10) γ γ P The series converges, for example, when |wα| > β6=α |wβ |. Moreover, 8α satisﬁes the differential equations

u v X X (∂ − ∂ )8α = 0 if (uγ − vγ ) = 0 and γ (uγ − vγ ) = 0 (11) γ γ and X wγ ∂γ 8α = 1, (12) γ X γ wγ ∂γ 8α = α. (13) γ AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 493

Proof Use the power series expansion of the logarithm function to write

m−1 β X (−1) X wβ z m log f (z)/zα = log w + . α m w zα m≥1 β6=α α Now

Q kβ k β (−1)m−1 w zβ m (−1)m−1 m! w z β X X β = X X β α Q m mα m wαz m kβ ! wα z m≥1 β6=α m≥1 6kβ =m (−k − 1)! X kα−1 α k 6kγ γ = (−1) Q w z . kβ ! Lα Here all sums and products indexed by β are taken over β ∈ C \{α} while γ ranges C P over all of C and Lα = {k ∈ Z ; kα < 0, kβ ≥ 0, kγ = 0}. The constant terms in this expression, considered as monomials in the z-variables, are precisely those corresponding to the set Kα, and we have proved (9). To verify the differential equations, we differentiate under the sign of integration defining 8α. By a simple computation, − u α X 6uγ γ 6uγ ∂ log f (z)/z = − uγ − 1 ! z − f (z) , P P which depends only on uγ and uγ γ . Also, γ X X wγ z w ∂ log f (z)/zα = = 1. γ γ f (z) This verifies (11) and (12). Finally, γ X α X γ j wγ z z j ∂ f/∂z j γ j w ∂ log f (z)/z = = . γ γ f (z) f (z) Comparing this to definition (2) of the order of a component proves relation (13).

Example = + n+1+· · ·+ n+1+ ··· Let us consider the polynomial f (z) 1 z1 zn az1 zn in n variables. The Newton polytope of this polynomial is a simplex, and the terms correspond to the vertices and the barycenter of the Newton polytope. The complement of the amoeba of f has at least n +1 components corresponding to the vertices of the Newton polytope, and in addition there may be a component corresponding to the single interior point (1,..., 1). For the case of n = 2, see Figure 2. Let Kn denote the set of a ∈ C such n that R \ A f does not have a component of order (1,..., 1). 494 PASSARE and RULLGARD˚

Figure 2. Amoebas, spines, and triangulated Newton polytopes of + 3 + 3 + = = − the polynomial 1 z1 z2 az1z2 for a 0 and a 6

PROPOSITION 2 = + n+1 + · · · + n+1 + ··· n \ Let f (z) 1 z1 zn az1 zn. Then R A f has a component of n order (1,..., 1) if and only if A f does not contain the origin. Hence R \ A f has such a component precisely if a 6∈ Kn = − t0 − · · · − tn; |t0| = · · · = |tn| = 1, t0 ··· tn = 1 .

Proof P α Let M be an invertible (n×n)-matrix with integer entries, and let f (z) = α aαz be P Mα any Laurent polynomial. Define a new Laurent polynomial M f = α aαz where a multi-index α is regarded as a column vector. It is not difficult to show that the linear mapping x 7→ M T x (where T denotes transpose) takes the amoeba of M f onto the amoeba of f . Now let f be the special polynomial in the proposition, and consider the M that map the Newton polytope of f onto a translate of itself. The set of such M can be identified with the symmetric group Sn+1 via its action on the vertices. Then M f coincides with f up to an invertible factor; hence they have the same amoeba. In T particular, M maps the component of order (1,..., 1) in the complement of A f (if it exists) onto itself. If x is any point in that component, then the convex hull of the T points M x, where M ranges over Sn+1, contains the origin. Since every component in the complement of the amoeba is convex, it follows that the component of order (1,..., 1) contains the origin. Conversely, if some component E contains the origin, then M T E also contains the origin and hence coincides with E. This implies that E has order (1,..., 1). AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 495

Now the amoeba of f contains the origin precisely if

= − + n+1 + · · · + n+1 ··· a (1 z1 zn )/z1 zn with |z1| = · · · = |zn| = 1 or, equivalently, a ∈ − t0 − · · · − tn; |t0| = · · · = |tn| = 1, t0 ··· tn = 1 .

We depict the sets Kn for n = 2 and n = 3 in Figure 3. Note that the cusps on the boundary correspond to polynomials deﬁning singular hypersurfaces.

Figure 3. The sets Kn for n = 2 and 3

Let us, in particular, consider the case of n = 2. The coefﬁcients cα are equal to zero when α is a vertex of the Newton polytope, while

X (3k − 1)! k−1 −3k c 1 1 = log |a| + Re (−1) a ( , ) (k!)3 k>0 when |a| > 3 by Theorem 2. In fact, it can easily be checked that the series converges even when |a| = 3. What happens with the spine when a approaches the boundary of K2 from the outside? For example, when a → −3, a numerical computation of the power series shows that c(1,1) converges to a limit with the approximate value 0.9693. This means that the complement of the spine has a rather large component that suddenly disappears when a enters K2.

5. Associated Monge-Ampere` measures Let f be a holomorphic function in Log−1(), where is a convex domain in Rn. We have seen that N f is a convex function that is linear in each component of \A f . Now we associate another object with f , namely, the Monge-Ampere` measure µ f of the Ronkin function. This is a positive measure with support on the amoeba A f . In Section 6 this measure is used to compute the maximal area of a plane amoeba. The goal of this section is to establish a few general results about µ f . First, we compute the total mass of µ f when f is a Laurent polynomial. Then we give an interpretation of µ f in terms of solutions to certain systems of equations. Let us remark 496 PASSARE and RULLGARD˚ that it might also be interesting to study the Monge-Ampere` measure of the Legendre transform of the Ronkin function. An investigation of such measures is not pursued in the present paper. If u is a smooth convex function, its Hessian Hess(u) is a positive semideﬁnite matrix that in a certain sense measures how convex u is near a given point. In particular, the determinant of the Hessian is a nonnegative function. The product of this function with ordinary Lebesgue measure is known as the real Monge-Ampere` measure of u, which we denote by Mu. In fact, the Monge-Ampere` operator can be extended as a continuous operator from the space of all convex functions with the topology of locally uniform convergence to the space of measures with the weak topology. In general, if u is a nonsmooth convex function, Mu will be a positive Borel measure (see [10]). We denote the Monge-Ampere` measure of N f by µ f . Hence µ f is a positive measure supported on the amoeba of f . It is interesting to consider also a generalization of the Monge-Ampere` operator. Notice that on smooth functions u, the operator M is actually the restriction to the di- ˜ agonal of a symmetric multilinear operator M on n functions u1,..., un. Conversely, M˜ can be recovered from M by the polarization formula

n 1 X X M˜ (u ,..., u ) = (−1)n−k M(u + · · · + u ). (14) 1 n n! j1 jk k=1 1≤ j1<···< jk ≤n

This expression still makes sense if u1,..., un are arbitrary convex functions, and by approximating u1,..., un with smooth convex functions it follows that ˜ M(u1,..., un) is a positive measure that depends multilinearly on the arguments u j . We call M˜ the mixed real Monge-Ampere` operator. The Monge-Ampere` operator can be given the following geometric interpretation. Let u be a convex function defined in a neighborhood of a Borel set E ⊂ Rn. Consider the set of ξ ∈ Rn such that the function hξ, xi − u(x) attains its global maximum at some x ∈ E. Then the Lebesgue measure of this set equals Mu(E). In the case of E = Rn, it is possible to give a similar characterization of the n mixed Monge-Ampere` operator. If u is a convex function defined in R , we let Ku denote the set of all ξ ∈ Rn such that the function hξ, xi − u(x) is bounded from above; in other words, the Legendre transform u˜(ξ) is finite.

PROPOSITION 3 If u, u1,..., un are convex functions, then the following statements hold: (i) Ku is a convex set; = + (ii) Ku1+u2 Ku1 Ku2 ; (iii) if Ku is bounded, then its volume equals the total mass of Mu; AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 497

(iv) if Ku1 ,..., Kun are all bounded, then their mixed volume equals the total mass ˜ of M(u1,..., un).

Proof + ⊂ Convexity of Ku and the inclusion Ku1 Ku2 Ku1+u2 are trivial. It is well known that the Legendre transform of u1 + u2 is the lower semicontinuous regularization of the function ξ 7→ inf u˜1(ξ1) +u ˜2(ξ2) ξ1+ξ2=ξ ∈ (see [6, Th. 2.2.5]). If ξ Ku1+u2 , it follows that there is a constant C such that there exist ξ1, ξ2 with u˜1(ξ1)+u ˜2(ξ2) ≤ C and ξ1 +ξ2 arbitrarily close to ξ. Since u˜1, u˜2 are bounded from below, we may assume (after possibly increasing C) that u˜ j (ξ j ) ≤ C. The set L j of ξ j satisfying this inequality is closed by lower semicontinuity of u˜ j and bounded because u j is bounded in a neighborhood of the origin. By a compactness argument, we therefore obtain the existence of points ξ j ∈ L j ⊂ Ku j with ξ1+ξ2 = ξ. = + This proves that Ku1+u2 Ku1 Ku2 . Let G denote the set of ξ such that hξ, xi − u(x) attains a maximal value. We claim that int Ku ⊂ G ⊂ Ku. The second inclusion is obvious; for if a real-valued function attains a minimal value, then it certainly is bounded from below. To prove the ﬁrst inclusion, take a point ξ in the interior of Ku, and choose points ξ0, . . . , ξn in Ku whose convex hull contains ξ in its interior. Thus we have u(x) − hξ j , xi ≥ C j for j = 0,..., n. It follows that u(x) − hξ, xi ≥ max C j − hξ j − ξ, xi . j

Since the right-hand side goes to ∞ when x → ∞, it follows that the left-hand side attains a minimal value. Hence ξ ∈ G. Since Ku is convex, we have Vol(int Ku) = Vol(Ku), and it follows that n Vol(Ku) = Vol(G) = Mu(R ). This proves (iii). It follows immediately that (iv) is true when u1 = · · · = un. Since both sides are multilinear and symmetric, it follows that equality holds in general.

Now we can compute the total mass of the Monge-Ampere` measure µ f for a Laurent polynomial f .

THEOREM 4 If f is a Laurent polynomial, then the total mass of µ f is equal to the volume of the Newton polytope of f . If f1,..., fn are Laurent polynomials, then the total mass of ˜ M(N f1 ,..., N fn ) is equal to the mixed volume of the Newton polytopes of f1,..., fn. 498 PASSARE and RULLGARD˚

Proof

We need only show that K N f is the Newton polytope of f . Let N denote the Newton polytope. If ξ is in N , then hξ, xi− N f (x) ≤ hξ, xi− S(x), where S(x) is deﬁned by (3) and (4), and this latter function is certainly bounded from above. Conversely, if ξ ∈ n h i h i is outside N , take v R so that ξ, v > supη∈N η, v and a vertex α of N where this supremum is attained. If x belongs to the complement component of order α, then x + tv is also in that component for all t > 0. Hence hξ, x + tvi − N f (x + tv) = hξ − α, x + tvi − cα → ∞ as t → ∞, so ξ is not in K N f .

Our next result depends on the following relations between the real and complex Monge-Ampere` operators, due to Rashkovskii [9]. Suppose that u1,..., un are smooth convex functions on the domain . Let U1,..., Un be plurisubharmonic func- −1 tions on Log () deﬁned by U j (z) = u j (Log z). Then Z Z ˜ c c n! M(u1,..., un) = dd U1 ∧ · · · ∧ dd Un, (15) E Log−1(E) c ¯ where d = (∂ − ∂)/2πi. This remains true if one of the functions U j , say, U1, is −1 allowed to be an arbitrary smooth plurisubharmonic function in Log (), u1 being deﬁned by 1 Z U (z) dz ··· dz = 1 1 n u1(x) n . (2πi) Log−1(x) z1 ··· zn More generally, if U1,..., Un are arbitrary smooth plurisubharmonic functions on Log−1(), and 1 Z U (z) dz ··· dz = j 1 n u j (x) n , (16) (2πi) Log−1(x) z1 ··· zn then Z Z Z ˜ c (1) c (n) n! M(u1,..., un) = dd U1(t z) ∧ · · · ∧ dd Un(t z) dλ(t). E T n2 Log−1(E) (17) n2 2 { = (k) ; | (k)| = = } Here T denotes the real n -dimensional torus t (t j ) t j 1, j, k 1,... n (k) = (k) (k) equipped with the usual normalized Haar measure λ, and each t (t1 ,..., tn ) acts on Cn by componentwise multiplication. These formulas can be checked by direct computation. A proof of (15) when u1 = · · · = un can be found in [9]. The general case follows by polarization since both sides are multilinear. Formula (17) follows by reversing the order of integration on the right. Since the inner integral is constant 2 along certain n-dimensional submanifolds of T n , it is actually possible to omit some of the variables in the outer integration. This also proves the generalized version of (15) with U1 an arbitrary smooth plurisubharmonic function. The following result follows from formula (17) and the Poincare-Lelong´ formula. AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 499

THEOREM 5 −1 Let f1,..., fn be holomorphic functions in Log (), and let E ⊂ be a Borel ! ˜ set. Then n M(N f1 ,..., N fn )(E) is equal to the average number of solutions in Log−1(E) to the system of equations ( j) ( j) = = f j (t1 z1,..., tn zn) 0, j 1,..., n, (18) = ( j) { ; | ( j)| = = } as t (tk ) ranges over the torus t tk 1, j, k 1,..., n .

Proof If U j are smooth plurisubharmonic functions that converge to log | f j |, then u j defined by (16) converge to N f j . By the general properties of the real Monge-Ampere` oper- ˜ ˜ ator, this implies that M(u1,..., un) converges weakly to M(N f1 ,..., N fn ). Also, c (1) c (n) dd U1(t z)∧· · ·∧dd Un(t z) converges weakly to the sum of point masses at the (1) (n) solutions of f1(t z) = · · · = fn(t z) = 0. Hence the theorem follows by passing to the limit in (17) if we show only that Z c (1) c (n) dd U1(t z) ∧ · · · ∧ dd Un(t z) Log−1(E) remains uniformly bounded as U j → log | f j |. Here we may assume that E is compact and that U j is of the form U j = ψ(log | f j |), where ψ is a convex function, constant near −∞, which will converge to the identity function. Let ft (z) = (1) (n) ( f1(t z), . . . , fn(t z)). Then ft (z) is a holomorphic function in z and t defined 2 for z in a neighborhood of Log−1(E) and t in a complex neighborhood of T n . Us- ing compactness arguments, it is not difficult to show that there exists a constant C −1 such that the number of solutions in Log (E) to the equation ft (z) = w is bounded n2 n c above by C for almost all t ∈ T and w ∈ C . Since η = dd ψ(log |w1|) ∧ · · · ∧ c n dd ψ(log |wn|) induces a positive measure on C with total mass 1, it follows that Z ∗ 0 ≤ ft η ≤ C Log−1(E) for almost all t, and this completes the proof.

Notice that an alternative proof of Theorem 4 is to use Theorem 5 and the Bernstein theorem (see [1]) on the number of solutions to a system of polynomial equations. On the other hand, the Bernstein theorem follows from Theorems 4 and 5 if it is assumed that the number of solutions is a constant depending only on the Newton polytopes. In fact, Theorem 5 can be regarded as a localized version of the Bernstein theorem which also gives some information about where the solutions are likely to be. Finally, we note that the method used to prove Theorem 5 can also be used to derive an interpretation of the Laplacian of the Ronkin function. 500 PASSARE and RULLGARD˚

THEOREM 6 Let E be any Borel set in , and let 1 denote the Laplace operator. Then Z Z n−1 (n − 1)! 1N f = ω , E Log−1(E)∩ f −1(0)

−2 −2 where ω = (|z1| dz¯1 ∧ dz1 + · · · + |zn| dz¯n ∧ dzn)/4πi.

Proof Since Hess(|x|2/2) is the identity matrix, it follows that

1u = nM˜ (u, |x|2/2,..., |x|2/2) for any convex function u. Now ω = ddc| Log z|2/2, and since ddc log | f | is equal to the current of integration along f −1(0), it follows from (15) that Z Z c n−1 n! 1N f = n dd log | f | ∧ ω E Log−1(E) Z = n ωn−1. Log−1(E)∩ f −1(0)

6. Monge-Ampere` measures for planar curves We now give some results on the Monge-Ampere` measure which are speciﬁc to the two-variable case. For n = 1, the situation is of course trivial since the amoeba is a discrete point set and µ f is a sum of point masses at these points. For n = 2, we have the following estimate on the Monge-Ampere` measure.

THEOREM 7 Let f be a holomorphic function in two variables deﬁned on a circular domain −1 −2 Log (). Then µ f is greater than or equal to π times the Lebesgue measure on A f .

Proof Let F denote the set of critical values of the mapping Log : f −1(0) → . We include in F the images of any singular points of f −1(0). Since F is a null set for the Lebesgue measure, it is sufﬁcient to prove the inequality in A f \ F. Take a small neighborhood U of some point in A f \F such that there exist smooth functions φk, ψk deﬁned in U with the property that

−1 −1 [ x1+iφk (x) x2+iψk (x) Log (U) ∩ f (0) = (e , e ); x = (x1, x2) ∈ U . k AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 501

We claim that 1 X ∂ψk/∂x1 ∂ψk/∂x2 Hess N f = ± . (19) 2π −∂φk/∂x1 −∂φk/∂x2 k

Differentiating the integral (1) deﬁning N f with respect to x1, we obtain ∂ N 1 Z ∂ f/∂z dz dz f = 1 1 2 Re 2 ∂x1 (2πi) Log−1(x) f (z)z2 Z 1 dz2 = n f (·, z2), x1 2πi log |z2|=x2 z2 1 Z 2π x2+iy2 = n f (·, e ), x1 dy2. 2π 0

Here n( f (·, z2), x1) is the number of zeros minus the number of poles of the function z1 7→ f (z1, z2) inside the disc {log |z1| < x1} provided that it is meromorphic in that · − · 0 domain. In general, n( f ( , z2), x1) n( f ( , z2), x1) is equal to the number of zeros in { 0 | | } 0 the annulus x1 < log z1 < x1 when x1 < x1. The integrand in the last integral is a piecewise constant function with jumps of magnitude 1 at y2 = ψk(x). It follows that −1 the gradient of ∂ N f /∂x1 is given by a sum of terms ±(2π) grad ψk. This proves the ﬁrst row of identity (19), up to sign changes. The correct sign of each term can be x +iy found by observing that n( f (·, e 2 2 ), x1) is increasing as a function of x1; hence 2 2 all the terms contributing to ∂ N f /∂x1 should be positive. A similar computation involving ∂ N f /∂x2 proves the second row. However, we have not yet shown that the choices of signs in the two rows are consistent. We now prove that all the terms on the right-hand side of (19) are symmetric, positive deﬁnite matrices with determinant equal to 1. Take a point x and an index k. x +iφ (x) x +iψ (x) Differentiating the expression f (e 1 k , e 2 k ) = 0 with respect to x1 and x2 yields the equations

∂ f ∂φk ∂ f ∂ψk z1 1 + i + z2 · i = 0, ∂z1 ∂x1 ∂z2 ∂x1 ∂ f ∂φk ∂ f ∂ψk z1 · i + z2 1 + i = 0. ∂z1 ∂x2 ∂z2 ∂x2

Writing a = z1∂ f/∂z1, b = z2∂ f/∂z2, these equations have the solution ∂ψ /∂x ∂ψ /∂x 1 |a|2 Re(ab¯ ) k 1 k 2 = 2 . −∂φk/∂x1 −∂φk/∂x2 Im(ab¯ ) Re(ab¯ ) |b| This matrix clearly has determinant 1. Changing the sign if Im(ab¯ ) < 0, we also have that the diagonal elements are positive, so the matrix is positive deﬁnite. Since we have already observed that the diagonal elements in the right-hand side of (19) are positive, it follows that they are positive deﬁnite with determinant equal to 1. 502 PASSARE and RULLGARD˚

The inequality now follows from the following lemma since Log−1(x) intersects −1 f (0) in at least two points for all x in A f \ F.

LEMMA 2 × + ≥ + + If√A1, A2 are 2 2 positive deﬁnite matrices, then det(A1 A2) det A1 det A2 2 det A1 det A2. Equality holds if and only if A1 and A2 are real multiples of one another.

Proof = a j b j To see this, write A j b c and apply the Cauchy-Schwarz inequality to the j j √ q 2 √ vectors (b j , a j c j − b ) to obtain b b + det A det A ≤ a a c c . Then it j 1 2 1 2 1 2√1 2 + − − = + − ≥ − follows that√ det(A1 A2) det A1 det A2 a1c2 c1a2 2b1b2 2 a1a2c1c2 ≥ 2 − 2 2b1b2 2 det A1 det A2. The conditions for equality are that (b1, a1c1 b1) is 2 − 2 proportional to (b2, a2c2 b2) and that (a1, c1) is proportional to (a2, c2), which clearly is equivalent to A1 being proportional to A2.

As an immediate consequence of Theorems 4 and 7, we have the following estimate.

COROLLARY 1 Let f be a Laurent polynomial in two variables. Then the area of the amoeba of f is not greater than π 2 times the area of the Newton polytope of f .

On the contrary, when n ≥ 3, the volume of the amoeba of a polynomial is almost always inﬁnite. We now give a few simple examples of polynomials for which the inequality in Theorem 7 becomes an equality, at least at most points in the amoeba. For a detailed account of the cases where equality holds, and the amoeba hence has maximal area, we refer to [8]. There it is shown that the polynomials with amoebas of maximal area are (up to certain changes of variables) precisely the real polynomials deﬁning a special kind of so-called Harnack curves in the real plane.

Example Consider the polynomial f (z1, z2) = 1 + z1 + z2. It is not difﬁcult to see that for this polynomial, equality holds in all the computations in the proof of Theorem 7. Hence 2 the area of A f is equal to π /2. This can also be established by a direct computation. x x x x The amoeba A f is bounded by the curves e 1 + e 2 = 1 and |e 1 − e 2 | = 1. For reasons of symmetry (cf. the proof of Prop. 2), the spine S f divides the amoeba into three parts with equal area. The spine consists of the negative halves of the coordinate axes and the positive diagonal x1 = x2 ≥ 0. We compute the area of the part of A f AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 503 in the third quadrant {x1 ≤ 0, x2 ≤ 0}. There the amoeba is bounded by the curve x x2 = log(1 − e 1 ); hence its area is ∞ ∞ Z 0 X Z 0 ekt X 1 π 2 − log(1 − et ) dt = dt = = . −∞ −∞ k k2 6 k=1 k=1

Closely related to this is the classical Euler dilogarithm function Li2, given for −1 ≤ x ≤ 1 by the two equivalent expressions

∞ n Z log |x| X x t Li2(x) = = − log 1 − sgn(x) e dt. n2 −∞ n=1

2 In particular, one has Li2(0) = 0 and Li2(1) = π /6. In our next example we have − use for the odd part Li2 of the dilogarithm, which admits the representations ∞ x2n−1 sgn(x) Z log |x| 1 + et − = X = Li2 (x) log t dt (2n − 1)2 2 −∞ 1 − e n=1

− = − = 2 and satisﬁes Li2 (0) 0 and Li2 (1) π /8.

Example (see Fig. 4) Here we let a be a real number and consider the polynomials f (z1, z2) = fa(z1, z2) = a + z1 + z2 + z1z2.

We want to compute the number of points in Log−1(x) ∩ f −1(0) for a given point n x x ∈ R . For points z in this set, it must hold that |a + z1| = |1 + z1|e 2 . Conversely, if this equation holds, then (z1,(a+z1)/(1+z1)) is in this set. If θ = arg z1, the equation can be rewritten 2ex1 (e2x2 −a) cos θ = a2 +e2x1 −e2x2 −e2x1+2x2 . From this it follows immediately that the amoeba of f is deﬁned by the inequality 4e2x1 (e2x2 − a)2 ≥ (a2 + e2x1 − e2x2 − e2x1+2x2 )2. We now consider separately the cases of a < 0 and a > 0. If a < 0, then the amoeba is bounded by the four curves

(1 ± ex1 )(1 ± ex2 ) = 1 − a. (20)

If x is in the interior of the amoeba, then Log−1(x)∩ f −1(0) has precisely two points, whereas for x in the boundary of the amoeba Log−1(x) ∩ f −1(0) consists of a single −2 point. By Theorem 7, µ f is greater than or equal to π times the Lebesgue measure on the amoeba of f . For x in the interior of A f , the sum (19) representing Hess(N f ) consists of two terms. Since f has real coefﬁcients, the curve f −1(0) is invariant under complex conjugation and it follows that (φ1, ψ1) = −(φ2, ψ2); hence the two 504 PASSARE and RULLGARD˚

Figure 4. The amoebas of a + z1 + z2 + z1z2 for a = −5 and a = 5 matrices are equal. This means that all inequalities in the proof of Theorem 7 actually become equalities. It follows also from Theorem 5 that µ f has no mass on the boundary of the amoeba. Hence the area of the amoeba is equal to π 2. If a > 0, the amoeba is still bounded by the curves defined by (20), but in this case a choice of the + sign in both factors gives an empty set. The equation with − signs in both factors defines a curve with two components bounding two complement components of the amoeba, while the curves defined by the two remaining equations intersect at a pinch in the middle of the amoeba. In this case too, Log−1(0)∩ f −1(0) has exactly two points for all x in the interior of the amoeba. However, if x = (log a, log a)/2, then Log−1(x) ∩ f −1(0) contains a real curve. For all other points x on the boundary of the amoeba, Log−1(x) intersects f −1(0) in exactly one point. It follows just as in −2 the case of a < 0 that µ fa is equal to π times the Lebesgue measure in the interior of the amoeba. However, A fa is strictly smaller than A f−a , so its area is smaller than 2 π . The remaining mass of µ fa , which must have the same total mass as µ f−a , resides as a point mass at the pinch (log a, log a)/2. We compute the size of this point mass in two different ways. For simplicity, we assume that a > 1.

First, we may compute the difference in area of the amoebas A fa and A f−a . Note that A f−a \ A fa consists of four regions. The lower right-hand part is bounded by the x x x x curves x2 = log(e 1 − a) − log(e 1 + 1) and x2 = log(e 1 − a) − log(e 1 − 1), where x1 ranges from log a to ∞. Hence its area is

Z ∞ ex1 + 1 log x dx1. log a e 1 − 1

The upper right-hand part is bounded above for 0 < x1 < ∞ by the curve x2 = x x log(e 1 + a) − log(e 1 − 1), and it is bounded below by the two curves x2 = log(a − x x x x e 1 ) − log(e 1 − 1) for 0 < x1 < (log a)/2 and x2 = log(e 1 + a) − log(e 1 + 1) for AMOEBAS, MONGE-AMPEREMEASURES,ANDTRIANGULATIONS` 505

(log a)/2 < x1 < ∞. Hence its area is

Z (log a)/2 a + ex1 Z ∞ ex1 + 1 + log x dx1 log x dx1 0 a − e 1 (log a)/2 e 1 − 1 Z log a et + 1 Z ∞ et + 1 = + log t dt log t dt. (log a)/2 e − 1 (log a)/2 e − 1 The two remaining parts are congruent to the ones described above. Hence the size of the point mass at (log a, log a)/2 is equal to

4 Z ∞ et + 1 8 1 dt = − √ 2 log t 2 Li2 . π (log a)/2 e − 1 π a Let us now present an alternative calculation of the same point mass by using instead our Theorem 5. Consider the system of equations

a + z1 + z2 + z1z2 = a + t1z1 + t2z2 + t1t2z1z2 = 0, (21) where t1, t2 are complex parameters with modulus 1. We want to determine, for all values of the parameters, the number of solutions to this system in −1 Log ((log a, log a)/2). Eliminating z2 from the system, we obtain the equation t t − z2 + at t − a − t + t z + a t − = 1( 2 1) 1 ( 1 2 1 √ 2) 1 ( 2 1) 0. This equation has either two roots on the circle |z1| = a or one root inside and one root outside this cir- √cle. In the former case, it is easy to see that the solutions for z2 also have modulus a. Now the two roots of the equation αζ 2 + βζ + γ = 0 have the same modulus precisely if β¯2(β2 − 4αγ ) is real and negative. Applying this to our equation, we ob- 2 2 2 tain the condition (at¯1t¯2 − a − t¯1 + t¯2) ((at1t2 − a − t1 + t2) − 4at1(t2 − 1) ) ≤ 2 0. Since (at¯1t¯2 − a − t¯1 + t¯2) /(t¯1t¯2) ≤ 0 for all t1, t2, this is equivalent to 2 2 i(φ+ψ) i(φ−ψ) t¯1t¯2((at1t2 −a−t1 +t2) −4at1(t2 −1) ) ≥ 0. Writing t1 = −e , t2 = −e , this may be reformulated as

(a cos φ + cos ψ)2 ≥ (a − 1)2. (22)

When φ, ψ range over the square |φ| + |ψ| < π, the parameters t1, t2 sweep out the torus minus a null set. Restricting φ, ψ to this square, it is clear that inequal- + ≥ − ity (22) will be satisﬁed√ precisely if a cos φ cos ψ a 1 or, equivalently, | sin(φ/2) | ≤ cos(ψ/2)/ a. To each choice of φ, ψ in the interior of this region, there correspond two solutions of the system (21) in Log−1((log a, log a)/2). Hence the average number of solutions is equal to

16 Z π/2 cos t √ dt 2 arcsin , π 0 a 506 PASSARE and RULLGARD˚ and by Theorem 5, the size of the point mass is half of this quantity. The fact that these two calculations do indeed yield the same result now amounts precisely to the identity

Z π/2 − = | | ≤ Li2 (x) arcsin(x cos t) dt, x 1. 0 To verify this identity directly, one can ﬁrst observe that both sides vanish for x = 0, and then one can√ perform explicit differentiations showing that their derivatives both equal x−1 log (1 + x)/(1 − x). In the particular case of a = 1, there is a factorization f (z) = (z1 + 1)(z2 + 1), and the amoeba consists of the two lines {x1 = 0} and {x2 = 0}. The Monge-Ampere` measure µ f then degenerates into a single point mass at the origin.

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Passare Matematiska institutionen, Stockholms universitet, SE-10691 Stockholm, Sweden; [email protected]

Rullgard˚ Matematiska institutionen, Stockholms universitet, SE-10691 Stockholm, Sweden; [email protected]